Validate if Tree is BST in DSA Typescript - Time & Space Complexity

Time Complexity: Validate if Tree is BST

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a tree follows the rules of a Binary Search Tree (BST).

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function isBST(node: TreeNode | null, min: number = -Infinity, max: number = Infinity): boolean {
  if (node === null) return true;
  if (node.val <= min || node.val >= max) return false;
  return isBST(node.left, min, node.val) && isBST(node.right, node.val, max);
}
    

This code checks if a binary tree is a BST by making sure each node's value fits between allowed minimum and maximum limits.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

As the tree grows, the function visits every node once to check its value.

Input Size (n)	Approx. Operations
10	About 10 checks
100	About 100 checks
1000	About 1000 checks

Pattern observation: The number of operations grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check the tree grows in a straight line with the number of nodes.

Common Mistake

[X] Wrong: "The function only checks some nodes, so it runs faster than O(n)."

[OK] Correct: The function must visit every node to be sure the whole tree follows BST rules, so it always checks all nodes.

Interview Connect

Understanding this helps you explain how to verify tree properties efficiently, a common skill in coding interviews.

Self-Check

"What if we used an in-order traversal to check the BST property instead? How would the time complexity change?"

Practice

1. What does it mean for a binary tree to be a Binary Search Tree (BST)?

easy

A. Every node has exactly two children

B. The tree is balanced with equal height on both sides

C. All nodes have the same value

D. Every node's left subtree has smaller values and right subtree has larger values

Solution

1. Step 1: Understand BST property

   A BST requires that for each node, all values in the left subtree are smaller and all values in the right subtree are larger.

2. Step 2: Compare options with definition

   Every node's left subtree has smaller values and right subtree has larger values matches the BST property exactly, while others describe different tree properties.

3. Final Answer:

   Every node's left subtree has smaller values and right subtree has larger values -> Option D

4. Quick Check:

   BST property = Every node's left subtree has smaller values and right subtree has larger values [OK]

Hint: Remember BST means left < node < right for all nodes [OK]

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Assuming all values are equal

2. Which TypeScript function signature correctly represents a function to check if a binary tree is a BST?

easy

A. function isBST(root: any): void

B. function isBST(root: TreeNode | null): boolean

C. function isBST(root: TreeNode): number

D. function isBST(root: number): boolean

Solution

1. Step 1: Identify correct parameter and return types

   The function should accept a tree node or null and return a boolean indicating if it is a BST.

2. Step 2: Match options with correct signature

   function isBST(root: TreeNode | null): boolean matches this: input is TreeNode or null, output is boolean. Others have wrong types or return types.

3. Final Answer:

   function isBST(root: TreeNode | null): boolean -> Option B

4. Quick Check:

   Correct signature = function isBST(root: TreeNode | null): boolean [OK]

Hint: Input is root node or null, output is true/false [OK]

Common Mistakes:

• Using wrong parameter type like number
• Returning number instead of boolean
• Using void return type

3. What will be the output of the following TypeScript code snippet?

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function isBST(root: TreeNode | null, min: number = -Infinity, max: number = Infinity): boolean {
  if (!root) return true;
  if (root.val <= min || root.val >= max) return false;
  return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);
}

const tree = new TreeNode(10, new TreeNode(5), new TreeNode(15, new TreeNode(6), new TreeNode(20)));
console.log(isBST(tree));

medium

A. SyntaxError

B. true

C. false

D. RuntimeError

Solution

1. Step 1: Analyze the tree structure

   The root is 10, left child 5 (valid), right child 15 with left child 6 and right child 20.

2. Step 2: Check BST property on right subtree

   Node 6 is in right subtree of 10 but less than 10, violating BST property. So function returns false.

3. Final Answer:

   false -> Option C

4. Quick Check:

   6 in right subtree < 10 = false [OK]

Hint: Check if left < node < right recursively [OK]

Common Mistakes:

• Ignoring min/max constraints in recursion
• Assuming all subtrees are BST without checks
• Confusing node values in subtrees

4. Identify the bug in this TypeScript function to check if a tree is BST:

function isBST(root: TreeNode | null): boolean {
  if (!root) return true;
  if (root.left && root.left.val > root.val) return false;
  if (root.right && root.right.val < root.val) return false;
  return isBST(root.left) && isBST(root.right);
}

medium

A. It does not check the entire subtree values against min/max limits

B. It uses wrong comparison operators for left and right children

C. It returns true for null nodes incorrectly

D. It does not handle empty tree input

Solution

1. Step 1: Understand the function checks

   The function only compares immediate children values with current node, not entire subtree ranges.

2. Step 2: Explain why this is a bug

   BST requires all nodes in left subtree < current node and all in right subtree > current node, not just direct children.

3. Final Answer:

   It does not check the entire subtree values against min/max limits -> Option A

4. Quick Check:

   Missing min/max range check = It does not check the entire subtree values against min/max limits [OK]

Hint: Check full subtree ranges, not just immediate children [OK]

Common Mistakes:

• Only comparing direct children values
• Ignoring recursive min/max constraints
• Assuming null nodes cause errors

5. Given a binary tree where nodes can have duplicate values, which modification to the BST validation logic allows duplicates only on the right subtree? Choose the correct condition to replace root.val <= min in the validation function:

function isBST(root: TreeNode | null, min: number = -Infinity, max: number = Infinity): boolean {
  if (!root) return true;
  if (root.val <= min || root.val >= max) return false;
  return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);
}

hard

A. Change to root.val < min to allow duplicates on right

B. Change to root.val <= min and root.val < max

C. Change to root.val < min and root.val <= max

D. Change to root.val <= min and root.val <= max

Solution

1. Step 1: Understand duplicates on right subtree rule

   Duplicates allowed only on right means left subtree nodes must be strictly less than current node, right subtree nodes greater than or equal.

2. Step 2: Adjust min/max comparisons accordingly

   Replace root.val <= min with root.val < min. This fails only if strictly less than min (allows equal to min on right recursion), keeping root.val >= max for strict upper bound.

3. Final Answer:

   Change to root.val < min to allow duplicates on right -> Option A

4. Quick Check:

   Duplicates right subtree = Change to root.val < min [OK]

Hint: Use strict less on left, less or equal on right for duplicates [OK]

Common Mistakes:

• Allowing duplicates on left subtree
• Using <= on both sides
• Not changing min/max checks